What is the range of sqrt(x^2-4) ?

The range of sqrt(x^2-4) is f(x)>= 0

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range of sqrt(x^2-4)

range

x^{2} - 4

f (x) \geq 0

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

x^{2} - 4 : x \leq - 2 or x \geq 2

Extreme Points of

x^{2} - 4 :

Minimum

(- 2, 0),

Minimum

(2, 0)

Find the range for the interval

- \infty < x \leq - 2 : 0 \leq f (x) < \infty

Find the range for the interval

2 \leq x < \infty : 0 \leq f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) \geq 0 or f (x) \geq 0

f (x) \geq 0

f (x) = \frac{x ^{2} - x - 56}{2 x - 16}

f (x) = (x - 3)^{\frac{2}{3}}

cos (x - \frac{π}{2})

f (x) = \frac{1}{1 - 1 - x}

f (x) = 3 x - 2 y = 12